$A$ hypothetical radioactive nucleus decays according to the following series:
$A \xrightarrow{\alpha} A_1 \xrightarrow{\beta^-} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4$
If the mass number and atomic number of $A$ are respectively $180$ and $72$,then the atomic number and mass number of $A_4$ will be respectively:

In the nuclear reaction $_{85}X^{297} \to Y + 4\alpha$,$Y$ is:

In a radioactive sample,${ }_{19}^{40} K$ nuclei decay into stable ${ }_{20}^{40} Ca$ nuclei with a decay constant of $4.5 \times 10^{-10} \text{ per year}$ or into stable ${ }_{18}^{40} Ar$ nuclei with a decay constant of $0.5 \times 10^{-10} \text{ per year}$. Given that in this sample,all the stable ${ }_{20}^{40} Ca$ and ${ }_{18}^{40} Ar$ nuclei are produced by the ${ }_{19}^{40} K$ nuclei only. In time $t \times 10^9 \text{ years}$,if the ratio of the sum of stable ${ }_{20}^{40} Ca$ and ${ }_{18}^{40} Ar$ nuclei to the radioactive ${ }_{19}^{40} K$ nuclei is $99$,the value of $t$ will be: [Given $\ln 10 = 2.3$]

$A$ radioactive substance emits

$A$ free neutron decays spontaneously into:

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
$^{38}S \xrightarrow{2.48 \ h} ^{38}Cl \xrightarrow{0.62 \ h} ^{38}Ar$
Assume that we start with $1000$ $^{38}S$ nuclei at time $t = 0$. The number of $^{38}Cl$ nuclei is zero at $t = 0$ and will again be zero at $t = \infty$. At what value of $t$ would the number of $^{38}Cl$ nuclei be a maximum?